solve -x+y=5,y=(x-3)^2+2 using substitution

Answer 1

To solve the system of equations given by

1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\)

using substitution, we can substitute the expression for \(y\) from the second equation into the first equation.

Starting with the first equation, we can express \(y\) as follows:

\[ y = x + 5 \]

Now we substitute this expression for \(y\) into the second equation:

\[ x + 5 = (x - 3)^2 + 2 \]

Next, we simplify the equation:

1. Expand \((x - 3)^2\): \[ (x - 3)^2 = x^2 - 6x + 9 \]

2. Substitute back: \[ x + 5 = x^2 - 6x + 9 + 2 \] \[ x + 5 = x^2 - 6x + 11 \]

Now, rearranging the equation to set it to zero:

\[ 0 = x^2 - 6x + 11 - x - 5 \] \[ 0 = x^2 - 7x + 6 \]

Now we can factor the quadratic equation:

\[ 0 = (x - 1)(x - 6) \]

Setting each factor to zero gives us:

1. \(x - 1 = 0 \Rightarrow x = 1\)
2. \(x - 6 = 0 \Rightarrow x = 6\)

Now that we have the values for \(x\), we can find the corresponding \(y\) values using the first equation \(y = x + 5\):

1. For \(x = 1\): \[ y = 1 + 5 = 6 \] So one solution is \((1, 6)\).

2. For \(x = 6\): \[ y = 6 + 5 = 11 \] So another solution is \((6, 11)\).

In conclusion, the solutions to the system of equations are:

\[ (1, 6) \quad \text{and} \quad (6, 11) \]